Search

Text Color

Margin Size

Font Type

Enable Dyslexic Font

11.4: Solving Equations of the Form ax = b and x/a = b

Last updated

Jun 8, 2024
Save as PDF
- 11.3: Solving Equations of the Form x + a = b and x - a = b
- 11.5: Applications I- Translating Words to Mathematical Symbols

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

Learning Objectives

be familiar with the multiplication/division property of equality
be able to solve equations of the form $ax = b$ and $\dfrac{x}{a} = b$
be able to use combined techniques to solve equations

Multiplication/ Division Property of Equality

Recall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side. From this, we can suggest the multiplication/division property of equality.

Multiplication/Division Property of Equality
Given any equation,

We can obtain an equivalent equation by multiplying both sides of the equation by the same nonzero number, that is, if $c \ne 0$ . then $a = b$ is equivalent to
$a \cdot c = b \cdot c$
We can obtain an equivalent equation by dividing both sides of the equation by the same nonzero number, that is, if $c \ne 0$ , then $a = b$ is equivalent to
$\dfrac{a}{c} = \dfrac{b}{c}$

The multiplication/division property of equality can be used to undo an association with a number that multiplies or divides the variable.

Sample Set A

Use the multiplication / division property of equality to solve each equation.

$6y = 54$

Solution

6 is associated with y by multiplication. Undo the association by dividing both sides by 6

$\begin{array} {rcl} {\dfrac{6y}{6}} & = & {\dfrac{54}{6}} \\ {\dfrac{\cancel{6} y}{\cancel{6}}} & = & {\dfrac{\begin{array} {c} {^9} \\ {\cancel{54}} \end{array}}{\cancel{6}}} \\ {y} & = & {9} \end{array}$

Check: When $y = 9$

$6y = 54$

becomes

$Does 6 times 9 equal 54? Yes.$

a true statement.

The solution to $6y = 54$ is $y = 9$ .

Sample Set A

$\dfrac{x}{-2} = 27$ .

Solution

-2 is associated with $x$ by division. Undo the association by multiplying both sides by -2.

$(-2) \dfrac{x}{-2} = (-2) 27$

$(\cancel{-2}) \dfrac{x}{\cancel{-2}} = (-2) 27$

$x = -54$

Check: When $x = -54$ .

$\dfrac{x}{-2} = 27$

becomes

$Does negative 54 over negative 2 equal 27? Yes.$

a true statement.

The solution to $\dfrac{x}{-2} = 27$ is $x = -54$

Sample Set A

$\dfrac{3a}{7} = 6$ .

Solution

We will examine two methods for solving equations such as this one.

Method 1: Use of dividing out common factors.

$\dfrac{3a}{7} = 6$

7 is associated with $a$ by division. Undo the association by multiplying both sides by 7.

$7 \cdot \dfrac{3a}{7} = 7 \cdot 6$

Divide out the 7's

$\cancel{7} \cdot \dfrac{3a}{\cancel{7}} = 42$

$3a = 42$

3 is associated with $a$ by multiplication. Undo the association by dviding both sides by 3.

$\dfrac{3a}{3} = \dfrac{42}{3}$

$\dfrac{\cancel{3} a}{\cancel{3}} = 14$

$a = 14$

Check: When $a = 14$ .

$\dfrac{3a}{7} = 6$

becomes

$Does the quantity 3 times 14, divided by 7 equal 6? Yes.$

a true statement.

The solution to $\dfrac{3a}{7} = 6$ is $a = 14$ .

Method 2: Use of reciprocals

Recall that if the product of two numbers is 1, the numbers are reciprocals. Thus $\dfrac{3}{7}$ and $\dfrac{7}{3}$ are reciprocals.

$\dfrac{3a}{7} = 6$

Multiply both sides of the equation by $\dfrac{7}{3}$ , the reciprocal of $\dfrac{3}{7}$ .

$\dfrac{7}{3} \cdot \dfrac{3a}{7} = \dfrac{7}{3} \cdot 6$

$\dfrac{\begin{array} {c} {^1} \\ {\cancel{7}} \end{array}}{\begin{array} {c} {\cancel{3}} \\ {^1} \end{array}} \cdot \dfrac{\begin{array} {c} {^1} \\ {\cancel{3} a} \end{array}}{\begin{array} {c} {\cancel{7}} \\ {^1} \end{array}} = \dfrac{7}{\begin{array} {c} {\cancel{3}} \\ {^1} \end{array}} \cdot \dfrac{\begin{array} {c} {^2} \\ {\cancel{6}} \end{array}}{1}$

$1 \cdot a = 14$

$a = 14$

Notice that we get the same solution using either method.

Sample Set A

$-8x = 24$ .

Solution

-8 is associated with $x$ by multiplication. Undo the association by dividing both sides by -8.

$\dfrac{-8x}{-8} = \dfrac{24}{-8}$

$x = -3$

Check: When $x = -3$ .

$-8x = 24$

becomes

$Does negative 8 times negative 3 equal 24? Yes.$

a true statement.

Sample Set A

$-x = 7$ .

Solution

Since $-x$ is actually $-1 \cdot x$ and $(-1)(-1) = 1$ . We can isolate $x$ by multiplying both sides of the equation by -1.

$(-1)(-x) = -1 \cdot 7$ .

$x = -7$

Check: When $x= 7$ .

$-x = 7$

becomes

$graphics5.png$

The solution to $-x = 7$ is $x = -7$ .

Practice Set A

Use the multiplication/division property of equality to solve each equation. Be sure to check each solution.

$7x = 21$

Answer: $x = 3$

Practice Set A

$-5x = 65$

Answer: $x = -13$

Practice Set A

$\dfrac{x}{4} = -8$

Answer: $x = -32$

Practice Set A

$\dfrac{3x}{8} = 6$

Answer: $x = 16$

Practice Set A

$-y = 3$

Answer: $y = -3$

Practice Set A

$-k = -2$

Answer: $k = 2$

Combining Techniques in Equation Solving

Having examined solving equations using the addition/subtraction and the multiplication/division principles of equality, we can combine these techniques to solve more complicated equations.

When beginning to solve an equation such as $6x - 4 = -16$ , it is helpful to know which property of equality to use first, addition/subtraction or multiplication/division. Recalling that in equation solving we are trying to isolate the variable (disassociate numbers from it), it is helpful to note the following.

To associate numbers and letters, we use the order of operations.

Multiply/divide
Add/subtract

To undo an association between numbers and letters, we use the order of operations in reverse.

Add/subtract
Multiply/divide

Sample Set B

Solve each equation. (In these example problems, we will not show the checks.)

$6x - 4 = -16$

Solution

-4 is associated with xx by subtraction. Undo the association by adding 4 to both sides.

$6x - 4 + 4 = -16 + 4$

$6x = -12$

6 is associated with $x$ by multiplication. Undo the association by dividing both sides by 6

$\dfrac{6x}{6} = \dfrac{-12}{6}$

$x = -2$

Sample Set B

$-8k + 3 = -45$

Solution

3 is associated with $k$ by addition. Undo the association by subtracting 3 from both sides.

$-8k + 3 - 3 = -45 - 3$

$-8k = -48$

-8 is associated with $k$ by multiplication. Undo the association by dividing both sides by -8.

$\dfrac{-8k}{-8} = \dfrac{-48}{-8}$

$k = 6$

Sample Set B

$5m - 6 - 4m = 4m - 8 + 3m$ .

Solution

Begin by solving this equation by combining like terms.

$m - 6 = 7m - 8$ Choose a side on which to isolate m. Since 7 is greater than 1, we'll isolate m on the right side.
Subtract m from both sides.

$-m - 6 - m = 7m - 8 - m$

$-6 = 6m - 8$

8 is associated with m by subtraction. Undo the association by adding 8 to both sides.

$-6 + 8 = 6m - 8 + 8$

$2 = 6m$

6 is associated with m by multiplication. Undo the association by dividing both sides by 6.

$\dfrac{2}{6} = \dfrac{6m}{6}$ Reduce,

$\dfrac{1}{3} = m$

Notice that if we had chosen to isolate m on the left side of the equation rather than the right side, we would have proceeded as follows:

$m - 6 = 7m - 8$

Subtract $7m$ from both sides.

$m - 6 - 7m = 7m - 8 - 7m$

$-6m - 6 = -8$

Add 6 to both sides,

$-6m - 6 + 6 = -8 + 6$

$-6m = -2$

Divide both sides by -6.

$\dfrac{-6m}{-6} = \dfrac{-2}{-6}$

$m = \dfrac{1}{3}$

This is the same result as with the previous approach.

Sample Set B

$\dfrac{8x}{7} = -2$

Solution

7 is associated with $x$ by division. Undo the association by multiplying both sides by 7.

$\cancel{7} \cdot \dfrac{8x}{\cancel{7}} = 7(-2)$

$7 \cdot \dfrac{8x}{7} = -14$

$8x = -14$

8 is associated with $x$ by multiplication. Undo the association by dividing both sides by 8.

$\dfrac{\cancel{8} x}{\cancel{8}} = \dfrac{-7}{4}$

$x = \dfrac{-7}{4}$

Practice Set B

Solve each equation. Be sure to check each solution.

$5m + 7 = -13$

Answer: $m = -4$

Practice Set B

$-3a - 6 = 9$

Answer: $a = -5$

Practice Set B

$2a + 10 - 3a = 9$

Answer: $a = 1$

Practice Set B

$11x - 4 - 13x = 4x + 14$

Answer: $x = -3$

Practice Set B

$-3m + 8 = -5m + 1$

Answer: $m = -\dfrac{7}{2}$

Practice Set B

$5y + 8y - 11 = -11$

Answer: $y = 0$

Exercises

Solve each equation. Be sure to check each result.

Exercise $\PageIndex{1}$

$7x = 42$

Answer: $x = 6$

Exercise $\PageIndex{2}$

$8x = 81$

Exercise $\PageIndex{3}$

$10 x = 120$

Answer: $x = 12$

Exercise $\PageIndex{4}$

$11 x = 121$

Exercise $\PageIndex{5}$

$-6a = 48$

Answer: $a = -8$

Exercise $\PageIndex{6}$

$-9y = 54$

Exercise $\PageIndex{7}$

$-3y = -42$

Answer: $y = 14$

Exercise $\PageIndex{8}$

$-5a = -105$

Exercise $\PageIndex{9}$

$2m = -62$

Answer: $m = -31$

Exercise $\PageIndex{10}$

$3m = -54$

Exercise $\PageIndex{11}$

$\dfrac{x}{4} = 7$

Answer: $x = 28$

Exercise $\PageIndex{12}$

$\dfrac{y}{3} = 11$

Exercise $\PageIndex{13}$

$\dfrac{-z}{6} = -14$

Answer: $z = 84$

Exercise $\PageIndex{14}$

$\dfrac{-w}{5} = 1$

Exercise $\PageIndex{1}$

$3m - 1 = -13$

Answer: $m = -4$

Exercise $\PageIndex{15}$

$4x + 7 = -17$

Exercise $\PageIndex{1}$

$2 + 9x = -7$

Answer: $x = -1$

Exercise $\PageIndex{16}$

$5 - 11x = 27$

Exercise $\PageIndex{17}$

$32 = 4y + 6$

Answer: $y = \dfrac{13}{2}$

Exercise $\PageIndex{18}$

$-5 + 4 = -8m + 1$

Exercise $\PageIndex{19}$

$3k + 6 = 5k + 10$

Answer: $k = -2$

Exercise $\PageIndex{20}$

$4a + 16 = 6a + 8a + 6$

Exercise $\PageIndex{21}$

$6x + 5 + 2x - 1 = 9x - 3x + 15$

Answer: $x = \dfrac{11}{2}$ or $5 \dfrac{1}{2}$

Exercise $\PageIndex{22}$

$-9y - 8 + 3y + 7 = -7y + 8y - 5y + 9$

Exercise $\PageIndex{23}$

$-3a = a + 5$

Answer: \(a = -\dfrac{5}{4})

Exercise $\PageIndex{24}$

$5b = -2b + 8b + 1$

Exercise $\PageIndex{25}$

$-3m + 2 - 8m - 4 = -14m + m - 4$

Answer: $m = -1$

Exercise $\PageIndex{26}$

$5a + 3 = 3$

Exercise $\PageIndex{27}$

$7x + 3x = 0$

Answer: $x = 0$

Exercise $\PageIndex{28}$

$7g + 4 - 11g = -4g + 1 + g$

Exercise $\PageIndex{29}$

$\dfrac{5a}{7} = 10$

Answer: $a = 14$

Exercise $\PageIndex{30}$

$\dfrac{2m}{9} = 4$

Exercise $\PageIndex{31}$

$\dfrac{3x}{4} = \dfrac{9}{2}$

Answer: $x = 6$

Exercise $\PageIndex{32}$

$\dfrac{8k}{3} = 32$

Exercise $\PageIndex{33}$

$\dfrac{3a}{8} - \dfrac{3}{2} = 0$

Answer: $a = 4$

Exercise $\PageIndex{34}$

$\dfrac{5m}{6} - \dfrac{25}{3} = 0$

Exercises for Review

Exercise $\PageIndex{35}$

Use the distributive property to compute $40 \cdot 28$

Answer: $40(30 - 2) = 1200 - 80 = 1120$

Exercise $\PageIndex{36}$

Approximating $\pi$ by 3.14, find the approximate circumference of the circle.

$A circle with radius 8cm.$

Exercise $\PageIndex{37}$

Find the area of the parallelogram.

$A parallelogram with base 20cm and height 11cm.$

Answer: 220 sq cm

Exercise $\PageIndex{38}$

Find the value of $\dfrac{-3(4 - 15) - 2}{-5}$

Exercise $\PageIndex{39}$

Solve the equation $x - 14 + 8 = -2$ .

Answer: $x = 4$

Learning Objectives

Multiplication/ Division Property of Equality

Sample Set A

Sample Set A

Sample Set A

Sample Set A

Sample Set A

Combining Techniques in Equation Solving

Sample Set B

Sample Set B

Sample Set B

Sample Set B

Exercises

Exercises for Review

Support Center

How can we help?